The product of two numbers is 2028, and their H.C.F is 13. The number of such pairs is: 1
Solution
Correct Answer: Option B
Let,
The two numbers be x and y
Here,
xy = 2028
Also 13 is their HCF, thus both numbers must be divisible by 13.
So,
let x = 13a and y = 13b,
then
13a × 13b = 2028
⇒ 169ab = 2028
⇒ ab = 2028/169
⇒ ab = 12
Therefore, required possible pair of values of x and y which are prime to each other are (1,12) and (3, 4).
Thus, the required numbers are (13,156) and (39, 52).
The number of possible pairs is 2.
The two numbers be x and y
Here,
xy = 2028
Also 13 is their HCF, thus both numbers must be divisible by 13.
So,
let x = 13a and y = 13b,
then
13a × 13b = 2028
⇒ 169ab = 2028
⇒ ab = 2028/169
⇒ ab = 12
Therefore, required possible pair of values of x and y which are prime to each other are (1,12) and (3, 4).
Thus, the required numbers are (13,156) and (39, 52).
The number of possible pairs is 2.
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